Thread: Another Probability.

1) Four friends, 2 females and 2 males are playing contract bridge. Partners are randomly assigned. What is the probability that two females will be partners for the first game?

I did:
P(2 females) = P(no 2 females)
= P(6C2)
(8C2)
= 15
28

and i got it wrong....dont know what i did wrong.

Just list the simple events, there aren't that many of them.

Hello, questiongirl!

This is easier/harder than we'd expect . . .

Four friends, 2 females and 2 males are playing contract bridge.
Partners are randomly assigned.
What is the probability that two females will be partners for the first game?

Suppose the four friends are: .


There are only three assignments for partners:

. .

Therefore: .


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Choose a female.
. . Either one, it makes no difference.

Now what is the probability that her partner is the other female?

There are 3 choices: .

Therefore: .

Thanks Soroban...you see i have misinterpret the question. i thought it means that there are 4 friends, 2females, 2males so in total their are 8people.
okay i get it. thank you!